Flux integral through tricky surface. Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Surface integral (stokes?)Help me understand a surface integral question?Evaluating the surface integral where S is part of a plane and inside a cylinderSurface integral of vector field over a unit ballFlux integral using Stokes' TheoremCalculate the flux through a closed surfaceSurface Integral problem. May be some misconception.Evaluating the surface integral over subset of Cylinder surfaceCalculating FluxShow that $ iint_S fracvec r^3 dS =0 $
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Flux integral through tricky surface.
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Surface integral (stokes?)Help me understand a surface integral question?Evaluating the surface integral where S is part of a plane and inside a cylinderSurface integral of vector field over a unit ballFlux integral using Stokes' TheoremCalculate the flux through a closed surfaceSurface Integral problem. May be some misconception.Evaluating the surface integral over subset of Cylinder surfaceCalculating FluxShow that $ iint_S fracvec rvec r dS =0 $
$begingroup$
Let $T$ be the area lying in the first octant where $xgeq0,ygeq0,zgeq0$ limited by the surfacs $z=a^2-x^2$ and $y=a^2-x^2$.
Calculate $iint_S vecFcdothatNdS$ where $vecF=(x,y,z)$ for $(x,y,z)inmathbbR^3$, $S$ is the part of the boundary $partial T$ to $T$ that lies on the surface $z=a^2-x^2$ and $hatN$ is the unit normal vector pointing out from $T$.
I have discussed the problem with a couple of friends, and we are all stuck. We where thinking about calculating $DivvecF$ but then again we struggle with the limits of that triple integral. Any help would be really great! Thanks in advance
calculus integration vector-fields divergence
$endgroup$
add a comment |
$begingroup$
Let $T$ be the area lying in the first octant where $xgeq0,ygeq0,zgeq0$ limited by the surfacs $z=a^2-x^2$ and $y=a^2-x^2$.
Calculate $iint_S vecFcdothatNdS$ where $vecF=(x,y,z)$ for $(x,y,z)inmathbbR^3$, $S$ is the part of the boundary $partial T$ to $T$ that lies on the surface $z=a^2-x^2$ and $hatN$ is the unit normal vector pointing out from $T$.
I have discussed the problem with a couple of friends, and we are all stuck. We where thinking about calculating $DivvecF$ but then again we struggle with the limits of that triple integral. Any help would be really great! Thanks in advance
calculus integration vector-fields divergence
$endgroup$
$begingroup$
Divergence theorem sounds like a good idea. Do you have a model answer to compare with? If so, what value do you expect?
$endgroup$
– StackTD
Apr 1 at 13:07
$begingroup$
@StackTD I do not have a model answer :/. We tried calculating the total divergence of $T$ and ended up with $8a^5/5$, but I reckon we have to subtract the flux through the part of the boundary which is not on $z=a^2-x^2$. But yeah, we're stuck. The limits we used for calculating total flux(div) was $0leq xleq a$, $0leq yleq a^2-x^2$ and $0leq zleq a^2-x^2$, with $DIVvecF=3$
$endgroup$
– JoeNTNU
Apr 1 at 13:16
$begingroup$
Is it $y=a^2-colorredx^2$, like in the comment before, or $y=a^2-colorredz^2$, as you wrote in the question?
$endgroup$
– StackTD
Apr 1 at 13:38
$begingroup$
@StackTD Oh, mistake by me. $y=a^2-x^2$, I edited it now.
$endgroup$
– JoeNTNU
Apr 1 at 13:42
$begingroup$
Alright, it was going to get a lot messier if that $x$ was in fact a $z$.
$endgroup$
– StackTD
Apr 1 at 14:41
add a comment |
$begingroup$
Let $T$ be the area lying in the first octant where $xgeq0,ygeq0,zgeq0$ limited by the surfacs $z=a^2-x^2$ and $y=a^2-x^2$.
Calculate $iint_S vecFcdothatNdS$ where $vecF=(x,y,z)$ for $(x,y,z)inmathbbR^3$, $S$ is the part of the boundary $partial T$ to $T$ that lies on the surface $z=a^2-x^2$ and $hatN$ is the unit normal vector pointing out from $T$.
I have discussed the problem with a couple of friends, and we are all stuck. We where thinking about calculating $DivvecF$ but then again we struggle with the limits of that triple integral. Any help would be really great! Thanks in advance
calculus integration vector-fields divergence
$endgroup$
Let $T$ be the area lying in the first octant where $xgeq0,ygeq0,zgeq0$ limited by the surfacs $z=a^2-x^2$ and $y=a^2-x^2$.
Calculate $iint_S vecFcdothatNdS$ where $vecF=(x,y,z)$ for $(x,y,z)inmathbbR^3$, $S$ is the part of the boundary $partial T$ to $T$ that lies on the surface $z=a^2-x^2$ and $hatN$ is the unit normal vector pointing out from $T$.
I have discussed the problem with a couple of friends, and we are all stuck. We where thinking about calculating $DivvecF$ but then again we struggle with the limits of that triple integral. Any help would be really great! Thanks in advance
calculus integration vector-fields divergence
calculus integration vector-fields divergence
edited Apr 1 at 13:42
JoeNTNU
asked Apr 1 at 12:51
JoeNTNUJoeNTNU
63
63
$begingroup$
Divergence theorem sounds like a good idea. Do you have a model answer to compare with? If so, what value do you expect?
$endgroup$
– StackTD
Apr 1 at 13:07
$begingroup$
@StackTD I do not have a model answer :/. We tried calculating the total divergence of $T$ and ended up with $8a^5/5$, but I reckon we have to subtract the flux through the part of the boundary which is not on $z=a^2-x^2$. But yeah, we're stuck. The limits we used for calculating total flux(div) was $0leq xleq a$, $0leq yleq a^2-x^2$ and $0leq zleq a^2-x^2$, with $DIVvecF=3$
$endgroup$
– JoeNTNU
Apr 1 at 13:16
$begingroup$
Is it $y=a^2-colorredx^2$, like in the comment before, or $y=a^2-colorredz^2$, as you wrote in the question?
$endgroup$
– StackTD
Apr 1 at 13:38
$begingroup$
@StackTD Oh, mistake by me. $y=a^2-x^2$, I edited it now.
$endgroup$
– JoeNTNU
Apr 1 at 13:42
$begingroup$
Alright, it was going to get a lot messier if that $x$ was in fact a $z$.
$endgroup$
– StackTD
Apr 1 at 14:41
add a comment |
$begingroup$
Divergence theorem sounds like a good idea. Do you have a model answer to compare with? If so, what value do you expect?
$endgroup$
– StackTD
Apr 1 at 13:07
$begingroup$
@StackTD I do not have a model answer :/. We tried calculating the total divergence of $T$ and ended up with $8a^5/5$, but I reckon we have to subtract the flux through the part of the boundary which is not on $z=a^2-x^2$. But yeah, we're stuck. The limits we used for calculating total flux(div) was $0leq xleq a$, $0leq yleq a^2-x^2$ and $0leq zleq a^2-x^2$, with $DIVvecF=3$
$endgroup$
– JoeNTNU
Apr 1 at 13:16
$begingroup$
Is it $y=a^2-colorredx^2$, like in the comment before, or $y=a^2-colorredz^2$, as you wrote in the question?
$endgroup$
– StackTD
Apr 1 at 13:38
$begingroup$
@StackTD Oh, mistake by me. $y=a^2-x^2$, I edited it now.
$endgroup$
– JoeNTNU
Apr 1 at 13:42
$begingroup$
Alright, it was going to get a lot messier if that $x$ was in fact a $z$.
$endgroup$
– StackTD
Apr 1 at 14:41
$begingroup$
Divergence theorem sounds like a good idea. Do you have a model answer to compare with? If so, what value do you expect?
$endgroup$
– StackTD
Apr 1 at 13:07
$begingroup$
Divergence theorem sounds like a good idea. Do you have a model answer to compare with? If so, what value do you expect?
$endgroup$
– StackTD
Apr 1 at 13:07
$begingroup$
@StackTD I do not have a model answer :/. We tried calculating the total divergence of $T$ and ended up with $8a^5/5$, but I reckon we have to subtract the flux through the part of the boundary which is not on $z=a^2-x^2$. But yeah, we're stuck. The limits we used for calculating total flux(div) was $0leq xleq a$, $0leq yleq a^2-x^2$ and $0leq zleq a^2-x^2$, with $DIVvecF=3$
$endgroup$
– JoeNTNU
Apr 1 at 13:16
$begingroup$
@StackTD I do not have a model answer :/. We tried calculating the total divergence of $T$ and ended up with $8a^5/5$, but I reckon we have to subtract the flux through the part of the boundary which is not on $z=a^2-x^2$. But yeah, we're stuck. The limits we used for calculating total flux(div) was $0leq xleq a$, $0leq yleq a^2-x^2$ and $0leq zleq a^2-x^2$, with $DIVvecF=3$
$endgroup$
– JoeNTNU
Apr 1 at 13:16
$begingroup$
Is it $y=a^2-colorredx^2$, like in the comment before, or $y=a^2-colorredz^2$, as you wrote in the question?
$endgroup$
– StackTD
Apr 1 at 13:38
$begingroup$
Is it $y=a^2-colorredx^2$, like in the comment before, or $y=a^2-colorredz^2$, as you wrote in the question?
$endgroup$
– StackTD
Apr 1 at 13:38
$begingroup$
@StackTD Oh, mistake by me. $y=a^2-x^2$, I edited it now.
$endgroup$
– JoeNTNU
Apr 1 at 13:42
$begingroup$
@StackTD Oh, mistake by me. $y=a^2-x^2$, I edited it now.
$endgroup$
– JoeNTNU
Apr 1 at 13:42
$begingroup$
Alright, it was going to get a lot messier if that $x$ was in fact a $z$.
$endgroup$
– StackTD
Apr 1 at 14:41
$begingroup$
Alright, it was going to get a lot messier if that $x$ was in fact a $z$.
$endgroup$
– StackTD
Apr 1 at 14:41
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
To make good use of the divergence theorem:
- verify that the surface integrals through the parts of the coordinate planes that, together with the two given surfaces, bound the region in the first octant, don't contribute to the flux (their share is $0$, which is easy to compute or even simply reason);
- note that due to symmetry in the two surfaces ($y leftrightarrow z$) that make up $partial T$ and with the given vector field, you expect the same flux through these two parts of $partial T$.
By the divergence theorem you find a total flux of $tfrac85a^5$ so the flux through the part $S$ is half of this: $tfrac45a^5$.

You can verify this by actually computing the surface integral. To do this, we parametrize $S$ (the paraboloid $z=a^2-x^2$ cut off by the paraboloid $y=a^2-x^2$) as follows:
$$vec r(u,v)=left(u,vleft(a^2-u^2right),a^2-u^2right) ;,; 0 le u le a;,; 0 le v le 1$$
Then:
$$vec N=fracpartial vec rpartial utimesfracpartial vec rpartial v=left(2a^2u-2u^3,0,4-u^2right)$$
and:
$$vec F cdot vec N=a^4-u^4$$
so:
$$iint_S vec F cdot vec N,mboxdS = int_0^1 int_0^a left(a^4-u^4right),mboxdu ,mboxdv=frac45a^5$$
$endgroup$
add a comment |
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1 Answer
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$begingroup$
To make good use of the divergence theorem:
- verify that the surface integrals through the parts of the coordinate planes that, together with the two given surfaces, bound the region in the first octant, don't contribute to the flux (their share is $0$, which is easy to compute or even simply reason);
- note that due to symmetry in the two surfaces ($y leftrightarrow z$) that make up $partial T$ and with the given vector field, you expect the same flux through these two parts of $partial T$.
By the divergence theorem you find a total flux of $tfrac85a^5$ so the flux through the part $S$ is half of this: $tfrac45a^5$.

You can verify this by actually computing the surface integral. To do this, we parametrize $S$ (the paraboloid $z=a^2-x^2$ cut off by the paraboloid $y=a^2-x^2$) as follows:
$$vec r(u,v)=left(u,vleft(a^2-u^2right),a^2-u^2right) ;,; 0 le u le a;,; 0 le v le 1$$
Then:
$$vec N=fracpartial vec rpartial utimesfracpartial vec rpartial v=left(2a^2u-2u^3,0,4-u^2right)$$
and:
$$vec F cdot vec N=a^4-u^4$$
so:
$$iint_S vec F cdot vec N,mboxdS = int_0^1 int_0^a left(a^4-u^4right),mboxdu ,mboxdv=frac45a^5$$
$endgroup$
add a comment |
$begingroup$
To make good use of the divergence theorem:
- verify that the surface integrals through the parts of the coordinate planes that, together with the two given surfaces, bound the region in the first octant, don't contribute to the flux (their share is $0$, which is easy to compute or even simply reason);
- note that due to symmetry in the two surfaces ($y leftrightarrow z$) that make up $partial T$ and with the given vector field, you expect the same flux through these two parts of $partial T$.
By the divergence theorem you find a total flux of $tfrac85a^5$ so the flux through the part $S$ is half of this: $tfrac45a^5$.

You can verify this by actually computing the surface integral. To do this, we parametrize $S$ (the paraboloid $z=a^2-x^2$ cut off by the paraboloid $y=a^2-x^2$) as follows:
$$vec r(u,v)=left(u,vleft(a^2-u^2right),a^2-u^2right) ;,; 0 le u le a;,; 0 le v le 1$$
Then:
$$vec N=fracpartial vec rpartial utimesfracpartial vec rpartial v=left(2a^2u-2u^3,0,4-u^2right)$$
and:
$$vec F cdot vec N=a^4-u^4$$
so:
$$iint_S vec F cdot vec N,mboxdS = int_0^1 int_0^a left(a^4-u^4right),mboxdu ,mboxdv=frac45a^5$$
$endgroup$
add a comment |
$begingroup$
To make good use of the divergence theorem:
- verify that the surface integrals through the parts of the coordinate planes that, together with the two given surfaces, bound the region in the first octant, don't contribute to the flux (their share is $0$, which is easy to compute or even simply reason);
- note that due to symmetry in the two surfaces ($y leftrightarrow z$) that make up $partial T$ and with the given vector field, you expect the same flux through these two parts of $partial T$.
By the divergence theorem you find a total flux of $tfrac85a^5$ so the flux through the part $S$ is half of this: $tfrac45a^5$.

You can verify this by actually computing the surface integral. To do this, we parametrize $S$ (the paraboloid $z=a^2-x^2$ cut off by the paraboloid $y=a^2-x^2$) as follows:
$$vec r(u,v)=left(u,vleft(a^2-u^2right),a^2-u^2right) ;,; 0 le u le a;,; 0 le v le 1$$
Then:
$$vec N=fracpartial vec rpartial utimesfracpartial vec rpartial v=left(2a^2u-2u^3,0,4-u^2right)$$
and:
$$vec F cdot vec N=a^4-u^4$$
so:
$$iint_S vec F cdot vec N,mboxdS = int_0^1 int_0^a left(a^4-u^4right),mboxdu ,mboxdv=frac45a^5$$
$endgroup$
To make good use of the divergence theorem:
- verify that the surface integrals through the parts of the coordinate planes that, together with the two given surfaces, bound the region in the first octant, don't contribute to the flux (their share is $0$, which is easy to compute or even simply reason);
- note that due to symmetry in the two surfaces ($y leftrightarrow z$) that make up $partial T$ and with the given vector field, you expect the same flux through these two parts of $partial T$.
By the divergence theorem you find a total flux of $tfrac85a^5$ so the flux through the part $S$ is half of this: $tfrac45a^5$.

You can verify this by actually computing the surface integral. To do this, we parametrize $S$ (the paraboloid $z=a^2-x^2$ cut off by the paraboloid $y=a^2-x^2$) as follows:
$$vec r(u,v)=left(u,vleft(a^2-u^2right),a^2-u^2right) ;,; 0 le u le a;,; 0 le v le 1$$
Then:
$$vec N=fracpartial vec rpartial utimesfracpartial vec rpartial v=left(2a^2u-2u^3,0,4-u^2right)$$
and:
$$vec F cdot vec N=a^4-u^4$$
so:
$$iint_S vec F cdot vec N,mboxdS = int_0^1 int_0^a left(a^4-u^4right),mboxdu ,mboxdv=frac45a^5$$
edited Apr 5 at 12:32
answered Apr 1 at 14:18
StackTDStackTD
24.3k2254
24.3k2254
add a comment |
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$begingroup$
Divergence theorem sounds like a good idea. Do you have a model answer to compare with? If so, what value do you expect?
$endgroup$
– StackTD
Apr 1 at 13:07
$begingroup$
@StackTD I do not have a model answer :/. We tried calculating the total divergence of $T$ and ended up with $8a^5/5$, but I reckon we have to subtract the flux through the part of the boundary which is not on $z=a^2-x^2$. But yeah, we're stuck. The limits we used for calculating total flux(div) was $0leq xleq a$, $0leq yleq a^2-x^2$ and $0leq zleq a^2-x^2$, with $DIVvecF=3$
$endgroup$
– JoeNTNU
Apr 1 at 13:16
$begingroup$
Is it $y=a^2-colorredx^2$, like in the comment before, or $y=a^2-colorredz^2$, as you wrote in the question?
$endgroup$
– StackTD
Apr 1 at 13:38
$begingroup$
@StackTD Oh, mistake by me. $y=a^2-x^2$, I edited it now.
$endgroup$
– JoeNTNU
Apr 1 at 13:42
$begingroup$
Alright, it was going to get a lot messier if that $x$ was in fact a $z$.
$endgroup$
– StackTD
Apr 1 at 14:41